Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 10010, M = 1e6 + 10;
int n, t, a[N], b[N], tot;
bool bo[N];
int pos[N];
int c[33], d[33], e[33], tmp[33];
std::pair<int, int> ans[M];
inline bool insert(int x, int id) {
  int y = 0;
  for (int i = 30; ~i; i--)
    if ((x >> i) & 1) {
      if (!c[i]) {
        c[i] = x;
        d[i] = y | (1 << i);
        e[i] = id;
        pos[id] = i;
        return true;
      }
      x ^= c[i];
      y ^= d[i];
    }
  return false;
}
inline int chk(int x) {
  int y = 0;
  for (int i = 30; ~i; i--)
    if ((x >> i) & 1) {
      if (!c[i]) return -1;
      x ^= c[i];
      y ^= d[i];
    }
  return y;
}
int main() {
  int i, j, z;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), bo[i] = insert(a[i], i);
  for (i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    if (chk(b[i]) == -1) return puts("-1"), 0;
  }
  for (i = 1; i <= n; i++)
    if (!bo[i]) {
      z = chk(a[i] ^ b[i]);
      for (j = 0; j <= 30; j++)
        if ((z >> j) & 1) ans[++tot] = std::make_pair(i, e[j]);
    }
  for (i = 1; i <= n; i++)
    if (bo[i]) {
      z = chk(a[i] ^ b[i]);
      if ((z >> pos[i]) & 1) {
        z ^= (1 << pos[i]);
        for (j = 0; j <= 30; j++)
          if (((z >> j) & 1) && e[j] > i) break;
        if (j > 30) continue;
        ans[++tot] = std::make_pair(e[j], i);
        ans[++tot] = std::make_pair(i, e[j]);
        ans[++tot] = std::make_pair(e[j], i);
        z ^= (1 << j);
        a[e[j]] = a[i];
        a[i] = b[i];
        for (j = 0; j <= 30; j++)
          if ((z >> j) & 1) ans[++tot] = std::make_pair(i, e[j]);
        for (j = 0; j <= 30; j++) tmp[j] = e[j], c[j] = d[j] = e[j] = 0;
        for (j = 0; j <= 30; j++)
          if (tmp[j]) insert(a[tmp[j]], tmp[j]);
      } else {
        for (j = 0; j <= 30; j++)
          if ((z >> j) & 1) ans[++tot] = std::make_pair(i, e[j]);
        a[i] = b[i];
        for (j = 0; j <= 30; j++) tmp[j] = e[j], c[j] = d[j] = e[j] = 0;
        for (j = 0; j <= 30; j++)
          if (tmp[j]) insert(a[tmp[j]], tmp[j]);
      }
      bo[i] = false;
    }
  for (i = n; i >= 1; i--)
    if (bo[i]) {
      ans[++tot] = std::make_pair(i, i);
      z = chk(b[i]);
      for (j = 0; j <= 30; j++)
        if ((z >> j) & 1) ans[++tot] = std::make_pair(i, e[j]);
    }
  printf("%d\n", tot);
  for (i = 1; i <= tot; i++) printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}